Question

我有下面的词典列表：

dictionary =[{'Flow': 100, 'Location': 'USA', 'Name': 'A1'},
            {'Flow': 90, 'Location': 'Europe', 'Name': 'B1'},
            {'Flow': 20, 'Location': 'USA', 'Name': 'A1'},
            {'Flow': 70, 'Location': 'Europe', 'Name': 'B1'}]

我想创建一个新的词典列表，其中所有Flow和Location相同的所有词典的Name值相加。我想要的输出将是：

new_dictionary =[{'Flow': 120, 'Location': 'USA', 'Name': 'A1'},
            {'Flow': 160, 'Location': 'Europe', 'Name': 'B1'},]

我该如何实现？

Answer 1

这是可能的，但在python中实现并非易事。我可以建议使用熊猫吗？使用import java.util.ArrayList; import java.util.HashMap; import java.util.Random; public class HashM { public static void main(String[] args) { HashMap<String, String> lyrics = new HashMap<>(); // KEY :- Lyrics AND Value is Album Name lyrics.put("that's me in the corner, that's me in the spotlight", "Losing My Religion - R.E.M."); lyrics.put("I will try not to breathe I can hold my head still with my hands at my knees", "Automatic for the People"); lyrics.put("Eu não aturo mais Sou um trator", "É Duda Brack"); randomLyrics(lyrics); } public static void randomLyrics(HashMap<String, String> lyrics) { ArrayList<String> lyricKey = new ArrayList<String>(lyrics.keySet()); if (lyrics.size() > 0) { Random rand = new Random(); int index = rand.nextInt(lyrics.size()); System.out.println("###::####Your Random Number is :-" + rand.nextInt(lyrics.size())); System.out.println(lyricKey.get(rand.nextInt(lyrics.size()))); } } }，groupby和sum很简单。

to_dict

要安装，请使用import pandas as pd (pd.DataFrame(dictionary) .groupby(['Location', 'Name'], as_index=False) .Flow.sum() .to_dict('r')) [{'Flow': 160, 'Location': 'Europe', 'Name': 'B1'}, {'Flow': 120, 'Location': 'USA', 'Name': 'A1'}]。

否则，您可以使用pip install --user pandas应用伪通用组操作。

itertools.groupby

Answer 2

虽然我也尽可能使用熊猫，但这是使用普通python的解决方案：

In [1]: import itertools

In [2]: dictionary =[{'Flow': 100, 'Location': 'USA', 'Name': 'A1'},
   ...:             {'Flow': 90, 'Location': 'Europe', 'Name': 'B1'},
   ...:             {'Flow': 20, 'Location': 'USA', 'Name': 'A1'},
   ...:             {'Flow': 70, 'Location': 'Europe', 'Name': 'B1'}]
   ...:

In [3]: import operator

In [4]: key = operator.itemgetter('Location', 'Name')

In [5]: [{'Flow': sum(x['Flow'] for x in g),
   ...:   'Location': k[0],
   ...:   'Name': k[1]}
   ...:  for k, g in itertools.groupby(sorted(dictionary, key=key), key=key)]
   ...:
   ...:
Out[5]:
[{'Flow': 160, 'Location': 'Europe', 'Name': 'B1'},
 {'Flow': 120, 'Location': 'USA', 'Name': 'A1'}]

另一种方法是使用defaultdict，这使您的表示形式稍有不同（尽管您可以根据需要将其转换回字典列表）：

In [11]: import collections

In [12]: cnt = collections.defaultdict(int)

In [13]: for r in dictionary:
    ...:     cnt[(r['Location'], r['Name'])] += r['Flow']
    ...:

In [14]: cnt
Out[14]: defaultdict(int, {('Europe', 'B1'): 160, ('USA', 'A1'): 120})

In [15]: [{'Flow': x, 'Location': k[0], 'Name': k[1]} for k, x in cnt.items()]
Out[15]:
[{'Flow': 120, 'Location': 'USA', 'Name': 'A1'},
 {'Flow': 160, 'Location': 'Europe', 'Name': 'B1'}]

Answer 3

不完全是您期望的输出，但是..

使用collections.Counter()

count = Counter()

for i in dictionary:
    count[i['Location'], i['Name']] += i['Flow']

print count

会给：

Counter({ ('Europe', 'B1'): 160, 
          ('USA', 'A1'): 120 })

我希望这至少会让您有所了解。

如果两个键值相同，如何对字典列表中的元素求和

3 个答案: